Question: Let $h(x)=-3x^4-18x^3$. For what values of $x$ does the graph of $h$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-6$ (Choice B) B $x=-3$ (Choice C) C $x=0$ (Choice D) D $h$ has no points of inflection.
Answer: We can find the inflection points of the graph of $h$ by looking for the intervals where its second derivative $h''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $h'$, we are analyzing $h''$. The second derivative of $h$ is $h''(x)=-36x(x+3)$. $h''(x)=0$ for $x=-3,0$. Since $h''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection points are $x=-3$ and $x=0$. Our possible inflection points divide the number line into three intervals: $\llap{-}7$ $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $x< \llap{-}3$ $\llap{-}3<x<0$ $x>0$ Let's evaluate $h''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h''(x)$ Verdict $x<-3$ $x=-4$ $h''(-4)=-144<0$ $h$ is concave down $\cap$ $-3<x<0$ $x=-1$ $h''(-1)=72>0$ $h$ is concave up $\cup$ $x>0$ $x=1$ $h''(1)=-144<0$ $h$ is concave down $\cap$ We can see that the graph of $h$ changes concavity at both $x=-3$ and $x=0$. In conclusion, these are the values of $x$ where the graph of $h$ has a point of inflection: $x=-3$ $x=0$